Electric fields + magnets = ouchies

[Poop-Loops]

Hi everyone, I just found this forum and am happy I did, being the only physics major in my physics 200 classes (community college :( ), I feel kind of isolated. Groups of people talking about computer science classes, engineering, and I'm alone, and my instructor/advisor still asks me everytime if I want to major in physics, even though I've said yes like 3 times now. It's just nice to have some people who have been there and done that to talk to. :)

Anyway, on to my question.

An electron has a velocity of 1.80 km/s (in the positive x direction) and an acceleration of 2.00 x 10e12 m/s2 (in the positive z direction) in uniform electric and magnetic fields. If the electric field has a magnitude of strength of 25.0 N/C (in the positive z direction), determine the following components of the magnetic field. (x,y,z)

Now, I don't want any numbers, just some hints on how to get started. I assume the magnetic field is going in an angle betweent he x and y axis, no z axis. I see the electric field, so I want to get rid of it by finding out its contribution, and attributing the exact opposite for magnetic field to get the 2e12, but I get a force to the 20th power, and an acceleration to the 50th power. I've just started magnetism, so I don't know if magnets that can do that are everyday things (I assume not :p), but it seems wrong.

Thanks in advance

PL

 

[AKG]

F = q(E + v x B)
ma = q(E + v x B)

You are given a, E, and v, and m and q are physical constants you can easily look up, so you just have to solve for B.

In case you don't get the above, F is the net force, which you know is mass times acceleration. But you know that the total force is simply the magnetic force and the electric force. The electric force is simply qE, and the magnetic force, you should know, is qv x B, so qE + qv x B = q(E + v x B). Now, since v x B is a cross product, and there's no inverse for it (i.e. you can't just divide away the v somehow to isolate B, you will have to write B as (Bx, By, Bz), compute the cross product of that with the given v, and you'll have a vector equation (which is really like three equations, since the x, y, and z components are independent), and three unknowns (Bx, By, and Bz). Solve for these, and you're done.

 

[Poop-Loops]

Cross Product?

I don't recall learning this. I definitely haven't done matrices.


If you mean this thing:

Cx = AyBz - AzBy
Cy = AzBx - AxBz
Cz = AxBy - AyBx

Then I don't understand, since I only have v in the x direction, and no B in the z direction. That only gives me Cz = AxBy.

PL

 

[Corneo]

Yea what you have there is correct for the 3 formulas.

 

[Poop-Loops]

...but I get the wrong answer...

PL

 

[HallsofIvy]

Cross Product?

I don't recall learning this. I definitely haven't done matrices.


If you mean this thing:

Cx = AyBz - AzBy
Cy = AzBx - AxBz
Cz = AxBy - AyBx

Then I don't understand, since I only have v in the x direction, and no B in the z direction. That only gives me Cz = AxBy.

PL

Yes, that is the cross product of the two vectors <Ax, Ay, Az> and <Bx, By, Bz>. If A is the electric field then you are told that A is the z direction (not x) so Ax= Ay= 0.
I don't see how you decide that there is "no B in the z direction". Bz is one of the things you want to find.

Since Ax= Ay= 0, A X B= C= <-AzBy, AzBx, 0>

 

[Poop-Loops]

Well, seeing as how the acceleration is in the z direction, I don't see how the magnetic field could be going there as well.

A is supposed to be velocity in the x dirction, B would be magnetic field.

So Ax = 1800m/s, Ay = Az = 0

PL

 

[Poop-Loops]

I hate to bump, but I'm really having problems with this well, problem. :)

PL

 

[Poop-Loops]

Ok, I made a breakthrough, and solved for Bz and By (finally understanding how this is supposed to work), but now I don't know how to find Bx, since Bx is always paired with a velocity that is 0.

PL

 

[Poop-Loops]

Oh, apparently you CAN'T determine it (or at least with my limited math knowledge), so "undetermined" is the answer.

Thanks for your help guys, I think I'll stick around here. This forum will probably help me get through a lot of tough times in my physics education career. :)

PL